Triple Cube

Find all integers between 100 and 1000 such that each digit is a different odd number and the sum of the cubes of the digits is the original three-digit number.
Source: 2/9/2009

If you try to construct 3-digit numbers with different odd digits 1, 3, 5, 7, and 9, you can make 5\times 4\times 3=60 3-digit numbers. For example, 135, 137, 139, etc. up to 971, 972, 975.

A long way to solve the problem is to go through the list of 60 numbers and calculate the cube of the digits and see if their sum is equal to the original number. For example,
1^3+3^3+5^3=153\neq 135
1^3+3^3+7^3=371\neq 137

A faster way is to calculate the sums of the cubes 1^3, 3^3, 5^3, 7^3, \textup{and }9^3 and see if the sums are equal to the original numbers.

How many possible sums are there? Since we are choosing 3 cubes from a set of 5 cubes and we do not care about order because addition is commutative, we have 10 sums as described below:

We can see right away that there are two solutions:

Answer: 153 and 371.


About mvtrinh

Retired high school math teacher.
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2 Responses to Triple Cube

  1. What a great web log. I spend hours on the net reading blogs, about tons of various subjects. I have to first of all give praise to whoever created your theme and second of all to you for writing what i can only describe as an fabulous article. I honestly believe there is a skill to writing articles that only very few posses and honestly you got it. The combining of demonstrative and upper-class content is by all odds super rare with the astronomic amount of blogs on the cyberspace.

    • mvtrinh says:

      Thank you for your generous comment. I try to explain the best I can so that readers understand what is going on. I tell my students that they should always ask three questions: Why? How? Can you explain?

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